SOLUTIONS

No problem is ever permanently closed. Any comments, new solutions, or new insights on old problems are always welcomed by the problem editor.

88. [1995, 140; 1996, 160–167] Proposed by Curtis Cooper and Robert E. Kennedy, Central Missouri State University, Warrensburg, Missouri.

Let m be a positive integer. Prove that

∏_{i=1}^{m} cos^{2} \frac{iπ}{2m + 1} = \frac{1}{4^{m}} .

Comment by Thomas C. Leong, The City College of City University of New York, New York, New York.

The discussion of Problem 232 in D. O. Shklarsky, N. N. Chentzov and I. M. Yaglom, The USSR Olympiad Problem Book, Dover Publications, New York, 1993, gives several solutions to show that

∏_{i=1}^{m} sin \frac{iπ}{2m + 1} = \frac{\sqrt{2m + 1}}{2^{m}}, ∏_{i=1}^{m−1} sin \frac{iπ}{2m} = \frac{\sqrt{m}}{2^{m−1}} ,

∏_{i=1}^{m} cos \frac{iπ}{2m + 1} = \frac{1}{2^{m}} , ∏_{i=1}^{m−1} cos \frac{iπ}{2m} = \frac{\sqrt{m}}{2^{m−1}} .

89. [1996, 36] Proposed by Stanley Rabinowitz, MathPro Press, Westford, Massachusetts.

Let ω be a primitive 49th root of unity. Prove that

∏_{{k=1 \atop gcd(k,49)=1}}^{49}(1 − ω^{k}) = 7.

Solution I by Lawrence Somer, The Catholic University of America, Washington, D.C.

We note that

∏_{{k=1 \atop gcd(k,49)=1}}^{49}(x − ω^{k}) = \frac{x^{49} − 1}{x^{7} − 1} = x^{42} + x^{35} + x^{28} + x^{21} + x^{14} + x^{7} + 1.

The first equality follows since the roots of

\frac{x^{49} − 1}{x^{7} − 1} = 0

are the 49th roots of unity which are not 7th roots of unity. Letting x = 1, we obtain that

∏_{{k=1 \atop gcd(k,49)=1}}^{49}(1 − ω^{k}) = 7.

Solution II by Joseph B. Dence, University of Missouri–St. Louis, St. Louis, Missouri.

Let

P = ∏_{{k=1 \atop gcd(k,49)=1}}^{49}(1 − ω^{k}).

The choice of primitive root is immaterial; choose

ω = e^{2πi/49}.

Then

ω^{k} = (ω^{49−k})^{∗},

where ^{∗} means complex conjugate, so

(1 − ω^{k})(1 − ω^{49−k}) = [1 − cos(2πk/49)]^{2} + sin^{2}(2πk/49) = 4 sin^{2}(πk/49).

Hence, by pairing off the factors in P, we obtain

P = \frac{4^{24} ∏_{k=1}^{24} sin^{2}(πk/49)}{4^{3} ∏_{k=7,14,21} sin^{2}(πk/49)} = 4^{21} \frac{∏_{k=1}^{24} sin^{2}(πk/49)}{∏_{k=1}^{3} sin^{2}(πk/7)}.

To evaluate the products we use the formula (L. B. W. Jolley, Summation of Series, 2nd ed., Dover, 1961, p. 190)

sin^{2}(π/n) sin^{2}(2π/n) ⋯ sin^{2}(((n − 1)/2)π/n) = n/2^{n−1},

where n is odd. Finally, we obtain

P = 2^{42} \frac{49/2^{48}}{7/2^{6}} = 7.

Solution III by Frank J. Flanigan, San Jose State University, San Jose, California.

We prove that if p is a prime, m is a natural number, and ω is a primitive p^{m}th root of unity, then

∏_{{k=1 \atop gcd(k,p^{m})=1}}^{p^{m}}(1 − ω^{k}) = p.

First, note that the left side here is f(1), where

f(x) = ∏_{{k=1 \atop gcd(k,p^{m})=1}}^{p^{m}}(x − ω^{k}).

Second, note that

f(x) = Φ_{p^{m}}(x) = the p^{m}th cyclotomic polynomial over the ring of integers.

This is because

{ω^{k} | 1 ≤ k ≤ p^{m}, gcd(k, p^{m}) = 1}

is precisely the set of primitive p^{m}th roots of unity in the complex numbers. Third, recall that

Φ_{p^{m}}(x) = Φ_{p}(x^{p^{m−1}}),

where Φ_{p}(x) is the pth cyclotomic polynomial. This follows from the fact that if ζ

is a primitive p^{m}th root of unity, then ζ^{p^{m−1}} is a primitive pth root of unity and hence, a zero of Φ_{p}(x).

Fourth, recall that

Φ_{p}(x) = x^{p−1} + x^{p−2} + ⋯ + x + 1.

Thus, the left side of the product is

f(1) = Φ_{p^{m}}(1) = Φ_{p}(1^{p^{m−1}}) = 1 + 1 + ⋯ + 1 + 1 = p,

which is what we asserted.

Solution IV by Kandasamy Muthuvel, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin.

The above result is a special case of the following proposition.

\underline{Proposition}. Let n > 1 be an integer and let ω be a primitive nth root of unity.

(1) If p is a prime divisor of n, then

∏_{{k=1 \atop gcd(k,p)=1}}^{n}(1 − ω^{k}) = p.

(2) If p is a prime divisor of n and n is not prime, then

∏_{{k=1 \atop gcd(k,p)=p}}^{n−1}(1 − ω^{k}) = n/p.

(3) If n is the product of two distinct primes, then

∏_{{k=1 \atop gcd(k,n)=1}}^{n−1}(1 − ω^{k}) = 1.

\underline{Proof}. Since

x^{n} − 1 = ∏_{k=1}^{n}(x − ω^{k}) = (x − 1) ∏_{k=1}^{n−1}(x − ω^{k}),

by differentiating both sides of the above equation with respect to x and by letting x = 1, we get

(∗)  n = ∏_{k=1}^{n−1}(1 − ω^{k}).

Let p be a prime divisor of n and let m = (n/p). Since

(ω^{p})^{k}, 1 ≤ k ≤ m

are solutions of the equation x^{m} − 1 = 0 and the solutions are distinct,

x^{m} − 1 = ∏_{k=1}^{m}(x − ω^{pk}).

Hence, following the method used in the beginning of the proof, we get

(∗∗)  (n/p) = m = ∏_{k=1}^{m−1}(1 − ω^{pk}) = ∏_{{k=1 \atop gcd(k,p)=p}}^{n−1}(1 − ω^{k}), when m > 1.

Part (1) of the proposition follows from (∗) when n = p and it follows from (∗) and (∗∗) when n is not prime.

Part (2) is in fact (∗∗).

To prove the third part of the proposition, let n = pq, where p and q are distinct primes. By (∗∗),

q = ∏_{k=1}^{q−1}(1 − ω^{pk}) and p = ∏_{k=1}^{p−1}(1 − ω^{qk}).

Clearly, pk ≠ qs for 1 ≤ k ≤ q − 1 and 1 ≤ s ≤ p − 1. Hence,

pq = ∏_{{k=1 \atop p|k or q|k}}^{n−1} (1 − ω^{k})

and by (∗) we get

pq = ∏_{k=1}^{n−1}(1 − ω^{k}).

These two equations imply that

1 = ∏_{{k=1 \atop gcd(k,n)=1}}^{n−1}(1 − ω^{k}).

Solution V by Thomas C. Leong, The City College of City University of New York, New York, New York.

In general, we show that if ω is a primitive nth root of unity, n ≥ 2, then

∏_{{k=1 \atop (k,n)=1}}^{n}(1 − ω^{k}) = \{ \begin{matrix}p, & if n = p^{α} is the power of a prime p, α ≥ 1 \\ 1, & otherwise. \end{matrix}

The primitive nth roots of unity are

{exp(2πir_{j}/n)}_{j=1}^{ϕ(n)},

where

{r_{j}}_{j=1}^{ϕ(n)}

is a reduced residue system modulo n. Now

{k : 1 ≤ k ≤ n, (k, n) = 1}

is a reduced residue system modulo n, and for any integer a relatively prime to n,

{ak : 1 ≤ k ≤ n, (k, n) = 1}

is also a reduced residue system modulo n. Thus, as k runs through the positive integers relatively prime to and not exceeding n, ω^{k} runs through the primitive nth roots of unity. Hence,

∏_{{k=1 \atop (k,n)=1}}^{n}(x − ω^{k})

is the nth cyclotomic polynomial

Φ_{n}(x).

Thus, we wish to show that for n ≥ 2,

(∗)  Φ_{n}(1) = \{ \begin{matrix}p, & if n = p^{α} is the power of a prime p, α ≥ 1 \\ 1, & otherwise. \end{matrix}

We induct on n. The case n = 2 is simple to verify, so suppose that (∗) holds for integers less than n. Now ζ is a root of x^{n} − 1 if and only if it is a primitive dth root of unity for some d which divides n; thus,

x^{n} − 1 = ∏_{d|n} Φ_{d}(x).

If n = p^{α} is a prime power, then

Φ_{p^{α}}(x) = \frac{x^{p^{α}} − 1}{∏_{{d|p^{α} \atop d≠p^{α}}} Φ_{d}(x)} = \frac{1 + x + x^{2} + ⋯ + x^{p^{α}−1}}{∏_{{d|p^{α} \atop d≠1,p^{α}}} Φ_{d}(x)} = \frac{1 + x + x^{2} + ⋯ + x^{p^{α}−1}}{Φ_{p}(x) · Φ_{p^{2}}(x) ⋯ Φ_{p^{α−1}}(x)}

and so by the inductive hypothesis

Φ_{p^{α}}(1) = \frac{p^{α}}{Φ_{p}(1) · Φ_{p^{2}}(1) ⋯ Φ_{p^{α−1}}(1)} = \frac{p^{α}}{p^{α−1}} = p.

If n is not a prime power, say,

n = p_{1}^{α_{1}}p_{2}^{α_{2}} ⋯ p_{r}^{α_{r}}

with r ≥ 2, α_{i} ≥ 1, then

Φ_{n}(x) = \frac{x^{n} − 1}{∏_{{d|n \atop d≠n}} Φ_{d}(x)} = \frac{1 + x + x^{2} + ⋯ + x^{n−1}}{∏_{{d|n \atop d≠1,n}} Φ_{d}(x)} .

By the inductive hypothesis, Φ_{d}(1) = 1 if d is not a prime power. Thus,

Φ_{n}(1) = \frac{n}{Φ_{p_{1}}(1) · Φ_{p_{1}^{2}}(1) ⋯ Φ_{p_{1}^{α_{1}}} (1) ⋯ Φ_{p_{r}}(1) · Φ_{p_{r}^{2}}(1) ⋯ Φ_{p_{r}^{αr}} (1)} = \frac{n}{p_{1}^{α_{1}}p_{2}^{α_{2}} · p_{r}^{α_{r}}} = 1

and we are done.

90. [1996, 36–37] Proposed by Joseph B. Dence, University of Missouri–St. Louis, St. Louis, Missouri.

The Fibonacci polynomials, {U_{n}(x)}, are defined by U_{1}(x) = 1, U_{2}(x) = x, and U_{n}(x) = xU_{n−1}(x) + U_{n−2}(x), for n ≥ 3.

(a) Derive a Binet–like formula for U_{n}(x).

(b) Prove that

(U_{n}(x))^{2} − U_{n−1}(x)U_{n+1}(x) = (−1)^{n−1}, n ≥ 2.

(c) Find a formula for the sum

∑_{k=1}^{n}(U_{k}(x))^{2}.

(d) Let {L_{n}} be the Lucas numbers: L_{1} = 1, L_{2} = 3, L_{n} = L_{n−1} + L_{n−2} (n ≥ 3). Prove that

U′_{n}(1) = \frac{nL_{n} − F_{n}}{5} ,

where F_{n} denotes the nth Fibonacci number and U′_{n}(x) denotes the derivative of U_{n}(x).

(e) Find a generating function for the U_{n}(x)’s, that is, a function f(x, y) such that, formally,

f(x, y) = ∑_{n=1}^{∞} U_{n}(x)y^{n}.

(f) Prove that for n > 1 all the zeroes of U_{n}(x) lie along the imaginary axis.

Solution I by the proposer.

(a) Suppose

U_{1}(x) = \frac{a(x) − b(x)}{d(x)}

and

U_{2}(x) = \frac{[a(x)]^{2} − [b(x)]^{2}}{d(x)} ,

so

U_{2}(x)/U_{1}(x) = x = a(x) + b(x).

Next,

U_{3}(x)/U_{1}(x) = x^{2} + 1 = [a(x)]^{2} + a(x)b(x) + [b(x)]^{2}.

Comparison then gives

[a(x)]^{2} + a(x)b(x) + [b(x)]^{2} = [a(x) + b(x)]^{2} + 1,

so a(x)b(x) = −1. It follows that

x = a(x) − \frac{1}{a(x)},

or

[a(x)]^{2} − xa(x) − 1 = 0,

and

a(x) = \frac{1}{2}(x ± \sqrt{x^{2} + 4}).

Let us choose the (+) sign in this expression. Then b(x) = x − a(x) and we conjecture that

U_{n}(x) = \frac{[\frac{1}{2}(x + \sqrt{x^{2} + 4})]^{n} − [\frac{1}{2}(x − \sqrt{x^{2} + 4})]^{n}}{\sqrt{x^{2} + 4}} .

This is automatically true for n = 1, 2; assume it to be true for n = k − 2, k − 1. Then,

U_{k}(x) = x(\frac{[a(x)]^{k−1} − [b(x)]^{k−1}}{d(x)} ) + (\frac{[a(x)]^{k−2} − [b(x)]^{k−2}}{d(x)} )

= \frac{[a(x)]^{k−2}(1 + xa(x)) − [b(x)]^{k−2}(1 + xb(x))}{d(x)}

= \frac{[a(x)]^{k−2}[a(x)]^{2} − [b(x)]^{k−2}[b(x)]^{2}}{d(x)}

identically. The conjecture is therefore true for all n by induction.

(b) This follows by straightforward induction, upon making use of a(x)b(x) = −1 and a(x) − b(x) = d(x). If we define U_{0}(x) = 0, then the relation is also true for n = 1.

(c) We have the identity

∑_{k=1}^{n} F_{k}^{2} = F_{n}F_{n+1}.

The analogous equation for U_{n}(x)’s would not be correct since the terms on the right–hand side would all be one power of x too great. Hence, we conjecture

∑_{j=1}^{n}[U_{j}(x)]^{2} = \frac{U_{n}(x)U_{n+1}(x)}{x}.

This holds for n = 1; assume it holds for n = k. Then

∑_{j=1}^{k+1}[U_{j}(x)]^{2} = \frac{U_{k}(x)U_{k+1}(x)}{x} + [U_{k+1}(x)]^{2}

= \frac{[U_{k+2}(x) − xU_{k+1}(x)]U_{k+1}(x)}{x} + [U_{k+1}(x)]^{2}

= \frac{U_{k+1}(x)U_{k+2}(x)}{x},

so the conjecture is true for all n by induction. (d) This follows by straightforward induction, upon making use of the identity

L_{j} = F_{j−1} + F_{j+1}.

(e) We have

yf(x, y) = ∑_{n=1}^{∞} U_{n}(x)y^{n+1} = ∑_{m=2}^{∞} U_{m−1}(x)y^{m}

xf(x, y) = ∑_{n=1}^{∞} xU_{n}(x)y^{n} = ∑_{n=2}^{∞} xU_{n}(x)y^{n} + xU_{1}(x)y

= ∑_{m=2}^{∞}[U_{m+1}(x) − U_{m−1}(x)]y^{m} + xyU_{1}(x)

\frac{1}{y} f(x, y) = ∑_{n=1}^{∞} U_{n}(x)y^{n−1} = ∑_{m=0}^{∞} U_{m+1}(x)y^{m}

= ∑_{m=2}^{∞} U_{m+1}(x)y^{m} + U_{1}(x) + U_{2}(x)y.

Combination of these equations yields

xf(x, y) − xyU_{1}(x) = \frac{1}{y} f(x, y) − U_{1}(x) − U_{2}(x)y − yf(x, y),

or equivalently,

f(x, y) = \frac{y}{1 − xy − y^{2}} .

(f) In the result of part (a) let x = 2i cos θ; then from deMoivre’s theorem we obtain

U_{n}(2i cos θ) = −i^{n+1}(\frac{sin nθ}{sin θ} ).

For θ satisfying 0 ≤ θ < 2π, the right–hand side is 0 only when θ = jπ/n, j = 1, 2, …, 2n − 1. But of these zeroes, only n − 1 are distinct because cos(jπ/n) = cos[2π − (jπ/n)]. If θ lies outside of [0, 2π) and is a zero of U_{n}, then θ is congruent modulo 2π to jπ/n, for some integer j ∈ [1, 2n − 1]. Hence, there are just n − 1 incongruent values of θ modulo 2π, and since U_{n}(x) is of degree n − 1, it can only have n − 1 zeroes. It follows that all the zeroes of U_{n}(x) are the pure imaginaries

{2i cos(jπ/n) : j = 1, 2, …, 2n − 1}.

We note that when n is even, but not odd, x = 0 is a zero (choose j = \frac{1}{2}n).

Solution II by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

(a) The characteristic equation for U_{n}(x) is λ^{2} − xλ − 1 = 0. The roots of this equation are u = u(x) = (x + \sqrt{x^{2} + 4})/2 and v = v(x) = (x − \sqrt{x^{2} + 4})/2. So,

U_{n}(x) = c_{1}u^{n} + c_{2}v^{n},

where c_{1} and c_{2} are constants. Using the initial conditions, one obtains the system of equations

\begin{aligned}U_{1}(x) & = c_{1}u + c_{2}v = 1, \\ U_{2}(x) & = c_{1}u^{2} + c_{2}v^{2} = x. \end{aligned}

The solution of this system is c_{1} = 1/\sqrt{x^{2} + 4}, c_{2} = −1/\sqrt{x^{2} + 4} if x ≠ ±2i. Hence, for x ≠ ±2i.

(1)  U_{n}(x) = (u^{n} − v^{n})/\sqrt{x^{2} + 4}.

(b) From (1),

[U_{n}(x)]^{2} − U_{n−1}(x)U_{n+1}(x)

= \frac{u^{2n} − 2(uv)^{n} + v^{2n} − [u^{2n} − u^{n−1}v^{n+1} − u^{n+1}v^{n−1} + v^{2n}]}{x^{2} + 4}

= \frac{2(−1)^{n−1} + (uv)^{n−1}(u^{2} + v^{2})}{x^{2} + 4}

= (−1)^{n−1} \frac{u^{2} + 2 + v^{2}}{x^{2} + 4}

= (−1)^{n−1} \frac{(u − v)^{2}}{x^{2} + 4}

= (−1)^{n−1}.

(c) The sequence of Fibonacci polynomials can be extended backwards by defining U_{0}(x) = 0. Then

∑_{k=1}^{n}[U_{k}(x)]^{2}

= ∑_{k=1}^{n}(U_{k}(x)[U_{k+1}(x) − U_{k−1}(x)]/x)

= \frac{1}{x}(∑_{k=1}^{n} U_{k}(x)U_{k+1}(x) − ∑_{k=1}^{n} U_{k}(x)U_{k−1}(x))

= \frac{1}{x}([U_{1}(x)U_{2}(x) − U_{1}(x)U_{0}(x)] + [U_{2}(x)U_{3}(x) − U_{2}(x)U_{1}(x)]

+ ⋯ + [U_{n}(x)U_{n+1}(x) − U_{n}(x)U_{n−1}(x)])

= \frac{U_{n}(x)U_{n+1}(x)}{x}.

(d) Using (1) it is straightforward to show that

U′_{n}(x) = \frac{\sqrt{x^{2} + 4}[nu^{n−1}u′ − nv^{n−1}v′] − xU_{n}(x)}{x^{2} + 4}

= \frac{n\sqrt{x^{2} + 4}[u^{n}/\sqrt{x^{2} + 4} + v^{n}/\sqrt{x^{2} + 4}] − xU_{n}(x)}{x^{2} + 4}

= \frac{n(u^{n} + v^{n}) − xU_{n}(x)}{x^{2} + 4}.

In particular, if x = 1, then

\begin{aligned}U′_{n}(1) & = \frac{n[u^{n}(1) + v^{n}(1)] − U_{n}(1)}{5} \\ & = \frac{nL_{n} − F_{n}}{5} . \end{aligned}

(e)

f(x, y) = ∑_{n=1}^{∞} U_{n}(x)y^{n}

= y + xy^{2} + ∑_{n=3}^{∞}[xU_{n−1}(x) + U_{n−2}(x)]y^{n}

= y + xy^{2} + xy ∑_{n=3}^{∞} U_{n−1}(x)y^{n−1} + y^{2} ∑_{n=3}^{∞} U_{n−2}(x)y^{n−2}.

Thus,

f(x, y) = y + xy^{2} + xy[f(x, y) − y] + y^{2}f(x, y).

So, (1 − xy − y^{2})f(x, y) = y and therefore, f(x, y) = y/(1 − xy − y^{2}).

(f) Using (1), U_{n}(x) = 0 is equivalent to u^{n} = v^{n} and so

\frac{u}{v} = 1^{1/n} = e^{2kπi/n}

for k = 0, 1, …, n − 1. Since

\frac{u}{v} = (x + \sqrt{x^{2} + 4})^{2}/(−4),

x + \sqrt{x^{2} + 4} = ±2ie^{kπi/n}.

Solving this equation for x gives

x = ±2i cos(\frac{kπ}{n} ).

91. [1996, 37] Proposed by Herta T. Freitag, Roanoke, Virginia.

Pythagoras did not have our computational facilities for trigonometric functions (calculators or tables) at his disposal, but he may have had a “feeling” for the aesthetic beauty of the golden ratio in his soul, as he is said to have chosen the pentagram as the design for the fraternity pin of his academy.

Let R, the radius of a circle be given. How could one obtain the area of the inscribed pentagram on this basis? (Leave your answer in terms of G = (\sqrt{5}+1)/2, the golden ratio.)

Solution I by the proposer.

Referring to Figure 1, the area of the pentagram is given by

(1)  K = 5(Area of △AME − Area of △AP E).

But,

Area of △AME = \frac{R^{2}}{2} sin 72^{∘}

and

Area of △AP E = \frac{a^{2}}{2} sin 72^{∘},

and thus,

(2)  K = \frac{5(R^{2} − a^{2})}{2} sin 72^{∘}.

For determining a, we consider △AMP (see Figure 2), and, using the law of sines

(3)  a = \frac{sin 36^{∘}}{sin 54^{∘}} R.

Now, we wish to obtain all these trigonometric functions. Referring to a regular decagon inscribed into our circle (see Figure 3) and letting its side be s_{10}, from similar triangles, we have

\frac{s_{10}}{R} = \frac{R − s_{10}}{s_{10}} ,

from which

s_{10} = R(\sqrt{5} − 1)/2 = \frac{R}{G}.

Furthermore, since

sin 18^{∘} = \frac{R}{2s_{10}} ,

(4)  sin 18^{∘} = \frac{1}{2G}.

Next, since sin 36^{∘} = 2 sin 18^{∘} cos 18^{∘},

(5)  sin 36^{∘} = \frac{\sqrt{4G + 3}}{2G^{2}}.

To obtain sin 54^{∘} = sin(3(18))^{∘}, we resort to sin 3x = sin x(3 − 4 sin^{2} x) to have

sin 54^{∘} = \frac{1}{2G}(3 − \frac{1}{G^{2}} ),

or

(6)  sin 54^{∘} = \frac{G}{2}.

To compute sin 72^{∘} = sin(4(18))^{∘}, we use

sin 4x = sin 2(2x) = 2 sin 2x(1 − 2 sin^{2} x).

Thus,

sin 72^{∘} = 2 sin 36^{∘}(1 − 2 sin^{2} 18^{∘})

or

sin 72^{∘} = \frac{\sqrt{4G + 3}}{G^{2}}(1 − \frac{1}{2G^{2}} ),

from which

(7)  sin 72^{∘} = \frac{\sqrt{4G + 3}}{2G}.

Finally, from (3), (5), and (6),

a = \frac{\sqrt{4G + 3}}{G^{3}R}

and

R^{2} − a^{2} = R^{2}(1 − \frac{4G + 3}{G^{6}}) = \frac{2R^{2}}{G^{3}} .

Hence, by (2),

K = \frac{5}{2} \frac{2R^{2}}{G^{3}} \frac{\sqrt{4G + 3}}{2G},

and the area of our pentagram becomes

K = \frac{5\sqrt{4G + 3}}{2G^{4}}R^{2}.

Figure 1.

Figure 2. Figure 3. 20

Solution II by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

Let x be the length of a side of the pentagon inscribed in the given circle. By the Law of Cosines,

\begin{aligned}x^{2} & = R^{2} + R^{2} − 2R^{2} cos 72^{∘} \\ & = 2R^{2}(1 − sin 18^{∘}). \end{aligned}

Since it is known that sin 18^{∘} = (G − 1)/2, x^{2} = R^{2}(3 − G) and so x = R\sqrt{3 − G}. Therefore, the area of the pentagon is given by

A_{1} = \frac{5}{4}x^{2} cot 36^{∘}.

Since cot 36^{∘} = G/\sqrt{3 − G},

A_{1} = \frac{5}{4}R^{2}G\sqrt{3 − G}.

Then the area, A, of the pentagram is A = A_{1} − 5A_{2}, where A_{2} is the area of the triangle in the figure below.

Then, A_{2} = xh/2. From the Law of Sines,

\frac{r}{sin 36^{∘}} = \frac{x}{sin 108^{∘}}

and so

r = \frac{x sin 36^{∘}}{sin 72^{∘}} = \frac{x sin 36^{∘}}{2 sin 36^{∘} cos 36^{∘}} = \frac{x}{2 sin 54^{∘}} = \frac{x}{G}.

Also,

r^{2} = (\frac{x}{2})^{2} + h^{2}

and hence, h = \sqrt{4 − G^{2}}x/(2G). Then

A_{2} = \sqrt{4 − G^{2}}x^{2}/(4G) = \sqrt{4 − G^{2}}R^{2}(3 − G)/(4G).

Therefore,

\begin{aligned}A & = \frac{5}{4}R^{2}G\sqrt{3 − G} − \frac{5R^{2}(3 − G)\sqrt{4 − G^{2}}}{4G} \\ & = \frac{5}{4}R^{2}(G\sqrt{3 − G} − \frac{(3 − G)\sqrt{4 − G^{2}}}{G} ). \end{aligned}

Solution III by Joseph B. Dence, University of Missouri–St. Louis, St. Louis, Missouri.

Pythagoras knew that intersecting diagonals cut each other in the golden ratio γ, that is,

\frac{\overline{MB}}{\overline{EM}} = γ.

Let \overline{AO} = R, \overline{OO′′} = h, \overline{DC} = s; as △DO′C ∼ △EO′B, we obtain

\frac{\overline{EB}}{\overline{DC}} = \frac{\overline{AD}}{\overline{DC}} = γ.

Then △OO′′D and △AO′′D furnish the Pythagorean equations R^{2} = (\frac{1}{2}s)^{2} + h^{2} and γ^{2}s^{2} = (\frac{1}{2}s)^{2} + (R + h)^{2}, respectively. These yield h = \frac{1}{2}γR, s = R\sqrt{3 − γ},

upon making use of \frac{1}{γ} = γ + 1. Additionally, \overline{AO′′} = R + h = R(1 + \frac{1}{2}γ) =

2\overline{AM′} + \overline{O′O′′} = 2\overline{AM′} + \frac{\overline{AM′}}{γ}, so

\overline{AM′} = \frac{1}{2}R(\frac{3γ + 1}{2γ + 1}).

Finally, we have

Area of pentagram = 2( Area of △AO′D + Area of △AMM′′)

= 2(\frac{1}{2}(\overline{AO′})(\overline{DO′′}) + \frac{1}{2}(\overline{MM′′})(\overline{AM′}))

= (2\overline{AM′})(\frac{1}{2}s) + (\frac{s}{1 + r} )(\overline{AM′})

= s(\overline{AM′})(1 + \frac{1}{1 + γ} )

= R\sqrt{3 − γ}(\frac{R}{2} \frac{3γ + 1}{2γ + 1})(\frac{γ + 2}{γ + 1})

= \frac{s}{2}R^{2}\sqrt{3 − γ}(\frac{2γ + 1}{5γ + 3})

= \frac{5}{2}R^{2}\sqrt{3 − γ}(2 − γ)

= \frac{5}{2}R^{2}\sqrt{18 − 11γ}.

Equivalent formulations of the final answer are possible.

92. [1996, 37; 1996, 89] Proposed by Joseph B. Dence, University of Missouri– St. Louis, St. Louis, Missouri.

It is easy to show that the homogeneous quadratic expression A_{n}^{2} − 2B_{n}^{2} is invariant for all members of the sequence \{γ_{n}\}_{n=1}^{∞}, defined by γ_{n} = (3 + 2\sqrt{2})^{n} = A_{n} +B_{n}\sqrt{2}. Find a homogeneous cubic expression that is invariant for all members of the sequence \{I_{n}\}_{n=1}^{∞}, defined by I_{n} = (1 + \sqrt[3]{2} + \sqrt[3]{4})^{n} = A_{n} + B_{n} \sqrt[3]{2} + C_{n} \sqrt[3]{4}.

Solution I by Lamarr Widmer, Messiah College, Grantham, Pennsylvania. We have

A_{n+1} + B_{n+1} \sqrt[3]{2} + C_{n+1} \sqrt[3]{4} = (1 + \sqrt[3]{2} + \sqrt[3]{4})(A_{n} + B_{n} \sqrt[3]{2} + C_{n} \sqrt[3]{4}). It is routine to expand the right hand side and equate coefficients of like terms to obtain A_{n+1} = A_{n} + 2B_{n} + 2C_{n}, B_{n+1} = A_{n} + B_{n} + 2C_{n}, and C_{n+1} = A_{n} + B_{n} + C_{n}. We will denote the general homogeneous cubic expression in three variables x, y, z as follows. Q(x, y, z) = ax^{3} + by^{3} + cz^{3} + dx^{2}y + exy^{2} + fx^{2}z + gxz^{2} + hy^{2}z + jyz^{2} + kxyz. Now the desired invariance condition can be stated as follows.

0 =Q(A_{n+1}, B_{n+1}, C_{n+1}) − Q(A_{n}, B_{n}, C_{n})
=Q(A_{n} + 2B_{n} + 2C_{n}, A_{n} + B_{n} + 2C_{n}, A_{n} + B_{n} + C_{n}) − Q(A_{n}, B_{n}, C_{n})
=A_{n}^{3}(b + c + d + e + f + g + h + j + k)
+ B_{n}^{3}(8a + c + 4d + 2e + 4f + 2g + h + j + 2k)
+ C_{n}^{3}(8a + 8b + 8d + 8e + 4f + 2g + 4h + 2j + 4k)
+ A_{n}^{2}B_{n}(6a + 3b + 3c + 4d + 4e + 5f + 4g + 3h + 3j + 4k)
+ A_{n}B_{n}^{2}(12a + 3b + 3c + 8d + 4e + 8f + 5g + 3h + 3j + 5k)

+ A_{n}^{2}C_{n}(6a + 6b + 3c + 6d + 6e + 4f + 4g + 5h + 4j + 5k)

+ A_{n}C_{n}^{2}(12a + 12b + 3c + 12d + 12e + 8f + 4g + 8h + 5j + 8k)

+ B_{n}^{2}C_{n}(24a + 6b + 3c + 16d + 10e + 12f + 6g + 4h + 4j + 8k)

+ B_{n}C_{n}^{2}(24a + 12b + 3c + 20d + 16e + 12f + 6g + 8h + 4j + 10k) + A_{n}B_{n}C_{n}(24a + 12b + 6c + 20d + 16e + 16f + 10g + 10h + 8j + 12k).

So we need to solve the system

b + c + d + e + f + g + h + j + k = 0

8a + c + 4d + 2e + 4f + 2g + h + j + 2k = 0 8a + 8b + 8d + 8e + 4f + 2g + 4h + 2j + 4k = 0 6a + 3b + 3c + 4d + 4e + 5f + 4g + 3h + 3j + 4k = 0 12a + 3b + 3c + 8d + 4e + 8f + 5g + 3h + 3j + 5k = 0 6a + 6b + 3c + 6d + 6e + 4f + 4g + 5h + 4j + 5k = 0 12a + 12b + 3c + 12d + 12e + 8f + 4g + 8h + 5j + 8k = 0 24a + 6b + 3c + 16d + 10e + 12f + 6g + 4h + 4j + 8k = 0 24a + 12b + 3c + 20d + 16e + 12f + 6g + 8h + 4j + 10k = 0 24a + 12b + 6c + 20d + 16e + 16f + 10g + 10h + 8j + 12k = 0.

This is a straightforward but sizable exercise (I used the Derive^{®} CAS) in row reduction. The solution is

a = 1, b = 2, c = 4, d = e = f = g = h = j = 0, k = −6.

The desired cubic expression is

A_{n}^{3} + 2B_{n}^{3} + 4C_{n}^{3} − 6A_{n}B_{n}C_{n}.

Solution II by the proposer.

Consider the number field ℚ(\sqrt[3]{2}); viewed as a vector space over ℚ, this has as a basis B = \{α_{1}, α_{2}, α_{3}\} = \{1, \sqrt[3]{2}, \sqrt[3]{4}\}. Let

α = ( \begin{matrix}α_{1} \\ α_{2} \\ α_{3} \end{matrix} );

then for any element γ ∈ ℚ(\sqrt[3]{2}), define the matrix A by

γα = Aα.

The norm of γ (in the field ℚ(\sqrt[3]{2}) is then given by (J. S. Chahal, Topics in Number Theory, Plenum Press, 1988, p. 75)

N(γ) = det A.

For γ = A_{n} + B_{n} \sqrt[3]{2} + C_{n} \sqrt[3]{4}, we obtain

N(γ) = | \begin{matrix}A_{n} & B_{n} & C_{n} \\ 2C_{n} & A_{n} & B_{n} \\ 2B_{n} & 2C_{n} & A_{n} \end{matrix} | = A_{n}^{3} + 2B_{n}^{3} + 4C_{n}^{3} − 6A_{n}B_{n}C_{n}.

This immediately gives N(I_{1}) = 1 + 2 + 4 − 6 = 1, so I_{1} is a unit in ℚ(\sqrt[3]{2}). Since norms in a number field obey N(x)N(y) = N(xy), it follows that N(I_{n}) = 1 for all n ∈ ℤ^{+}, and the cubic above is the desired homogeneous invariant expression. æ